Question

The value of integral $\int {\dfrac{{d\theta }}{{{{\cos }^3}\theta \sqrt {\sin 2\theta } }}}$ can be expressed as irrational function of $\tan \theta$ as:-
A. $\dfrac{{\sqrt 2 }}{5}\left( {\sqrt {{{\tan }^2}\theta + 5} } \right)\tan \theta + c$
B. $\dfrac{2}{5}\left( {{{\tan }^2}\theta + 5} \right)\sqrt {\tan \theta } + c$
C. $\dfrac{{\sqrt 2 }}{5}\left( {{{\tan }^2}\theta + 5} \right)\sqrt {\tan \theta } + c$
D. $\sqrt {\dfrac{2}{5}} \left( {{{\tan }^2}\theta + 5} \right)\sqrt {\tan \theta } + c$

Answer 1

We will simplify the given expression using the trigonometric formulas, $\sin 2\theta = 2\sin \theta \cos \theta$, $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\cos \theta = \dfrac{1}{{\sec \theta }}$. Then, substitute $\tan \theta = t$ in the given expression. At last, integrate the simplified expression using the formula, $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c$.

Complete step-by-step answer:
We have to find the value of $\int {\dfrac{{d\theta }}{{{{\cos }^3}\theta \sqrt {\sin 2\theta } }}}$
We know that $\sin 2\theta = 2\sin \theta \cos \theta$
$\int {\dfrac{{d\theta }}{{{{\cos }^3}\theta \sqrt {2\sin \theta \cos \theta } }}}$
Multiply and divide the expression by $\cos \theta$ and simplify it further.
$\int {\dfrac{{d\theta }}{{{{\cos }^4}\theta \dfrac{{\sqrt {2\sin \theta \cos \theta } }}{{\cos \theta }}}}} \\\ \Rightarrow \int {\dfrac{{d\theta }}{{{{\cos }^4}\theta \sqrt {\dfrac{{2\sin \theta \cos \theta }}{{{{\cos }^2}\theta }}} }}} \\\ \Rightarrow \int {\dfrac{{d\theta }}{{{{\cos }^4}\theta \sqrt {\dfrac{{2\sin \theta }}{{\cos \theta }}} }}} \\\$
Also, $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
$\Rightarrow \int {\dfrac{{d\theta }}{{{{\cos }^4}\theta \sqrt {2\tan \theta } }}}$
We also know that $\cos \theta = \dfrac{1}{{\sec \theta }}$
Hence,
$\Rightarrow \int {\dfrac{{{{\sec }^4}\theta d\theta }}{{\sqrt {2\tan \theta } }}} \\\ \Rightarrow \int {\dfrac{{{{\sec }^2}\theta \left( {{{\sec }^2}\theta } \right)d\theta }}{{\sqrt {2\tan \theta } }}} \\\$
Now, ${\sec ^2}\theta = 1 - {\tan ^2}\theta$
$\int {\dfrac{{{{\sec }^2}\theta \left( {1 - {{\tan }^2}\theta } \right)d\theta }}{{\sqrt {2\tan \theta } }}}$
We will solve the integration by substitution. We will substitute $\tan \theta = t$, then ${\sec ^2}\theta d\theta = dt$
$\int {\dfrac{{\left( {1 - {t^2}} \right)dt}}{{\sqrt {2t} }}} \\\ \Rightarrow \int {\left( {\dfrac{{dt}}{{\sqrt {2t} }} - \dfrac{{{t^2}dt}}{{\sqrt {2t} }}} \right)} \\\$
Now, we know that $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c$, therefore, the above integration can be simplified as,
$\Rightarrow \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{{{{\left( t \right)}^{ - \dfrac{1}{2} + 1}}}}{{\left( { - \dfrac{1}{2} + 1} \right)}} + \dfrac{{{t^{\dfrac{5}{2}}}}}{{\dfrac{5}{2}}}} \right) + c \\\ \Rightarrow \dfrac{1}{{\sqrt 2 }}\left( {2\sqrt t + \dfrac{{2{t^2}\sqrt t }}{5}} \right) + c \\\$
Take $2\sqrt t$ common from the bracket,
$\sqrt {2t} \left( {1 + \dfrac{{{t^2}}}{5}} \right) + c$
Substitute back the value of $t = \tan \theta$
Hence, the value of the integration is
$\sqrt {2\tan \theta } \left( {1 + \dfrac{{{{\tan }^2}\theta }}{5}} \right) + c \\\ \Rightarrow \dfrac{{\sqrt 2 }}{5}\sqrt {\tan \theta } \left( {5 + {{\tan }^2}\theta } \right) + c \\\$
Thus, option C is correct.

Note: Students must know how to simplify expression using trigonometric identities. Also, the expression given in the question is indefinite integral. But, if the integration has upper and lower limits, then the integral is known as a definite integral.